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Middle School Math Quiz

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Post by indophile Thu Jul 07, 2011 2:41 pm

Again questions from a math magazine.

1. Three pennies showing 2 heads and 1 tail lie on a table. Two of the coins are selected at random and turned over. What is the probability that now all three coins show the same side?

2. Anna and Bob play a game in which Anna begins by rolling a fair die, after which Bob tosses a fair coin. They take turns until one of them wins. Anna wins when she rolls a 6; Bob wins when the coin lands heads. What is the probability that Anna wins the game?

3. What is the average of all multiples of 7 between 7 and 777, inclusive? Please show two different ways of solving this problem, by using a simple formula, and by using simple commonsense.

4. The odometer of a family car shows 15,951 miles. The driver noticed that this number is palindromic (read the same forward or backward). Surprised, the driver saw his 3rd palindromic odometer reading (not counting 15,951) exactly 5 hours later. What is the car’s speed in mph over those 5 hours?

5. The fifth term of a geometric series is 5! And the sixth term is 6!. What is the fourth term?

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Post by Hellsangel Thu Jul 07, 2011 3:12 pm

1. 1/3

2. 1/12

3. (7+7*2+7*3..+7*111)/111 = (7*(1+2+3+...111))/111 = 7*112/2 = 392

common sense
(7+777)/2 = 392

4. 1st is 16061, 2nd palindromic # is 16161, 3rd one is 16261 - so 16261-15951/5 = 62mph

5. 20
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Post by indophile Thu Jul 07, 2011 3:23 pm

Hellsangel wrote:

The answer posted for #2, 1/12 is incorrect.

Others are fine.

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Post by charvaka Thu Jul 07, 2011 3:32 pm

For 2... in any round, Bob has 1 in 2 odds of winning while Anna only 1 in 6. So Bob is thrice as likely to win as Anna. One of them must win or the game won't stop. So Anna's probability is 1/4.
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Post by Hellsangel Thu Jul 07, 2011 3:34 pm

charvaka wrote:For 2... in any round, Bob has 1 in 2 odds of winning while Anna only 1 in 6. So Bob is thrice as likely to win as Anna. One of them must win or the game won't stop. So Anna's probability is 1/4.
What about a 'draw'?
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Post by charvaka Thu Jul 07, 2011 3:34 pm

I assumed, if it's a draw, they play another round.
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Post by indophile Thu Jul 07, 2011 3:39 pm

charvaka wrote:For 2... in any round, Bob has 1 in 2 odds of winning while Anna only 1 in 6. So Bob is thrice as likely to win as Anna. One of them must win or the game won't stop. So Anna's probability is 1/4.

Sorry, incorrect answer.

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Post by harharmahadev Thu Jul 07, 2011 3:48 pm

indophile wrote:Again questions from a math magazine.

1. Three pennies showing 2 heads and 1 tail lie on a table. Two of the coins are selected at random and turned over. What is the probability that now all three coins show the same side?



1/3

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Post by indophile Thu Jul 07, 2011 3:53 pm

To harharmahadev - Your answer, 1/3 for # 1 ia correct.

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Post by Hellsangel Thu Jul 07, 2011 3:56 pm

charvaka wrote:I assumed, if it's a draw, they play another round.
There are 12 outcomes
1 - H
1 - T
2 - H
2 - T
3 - H
3 - T
4 - H
4 - T
5 - H
5 - T
6 - H
6 - T
of these only 6 - T is the winning option for Anna
So isn't it 1/12?
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Post by indophile Thu Jul 07, 2011 3:58 pm

charvaka wrote:I assumed, if it's a draw, they play another round.

Right, like the question says - "they take turns until someone wins."

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Post by harharmahadev Thu Jul 07, 2011 4:02 pm

indophile wrote:Again questions from a math magazine.


2. Anna and Bob play a game in which Anna begins by rolling a fair die, after which Bob tosses a fair coin. They take turns until one of them wins. Anna wins when she rolls a 6; Bob wins when the coin lands heads. What is the probability that Anna wins the game?




If Anna and Bob were playing some strip-a-piece-of-clothing off with this game, then Anna is going to be buck naked well before Bob....even considering the fact that females have a lot more clothes than men.

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Post by charvaka Thu Jul 07, 2011 4:07 pm

That's if they play only one round... if Anna doesn't win round 1 and Bob doesn't win it either (6-H), then they play round 2.
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Post by Mosquito Thu Jul 07, 2011 4:14 pm

Probabilty of her winning 2/7?

1/6+5/6*1/2*1/6+5/6*1/2*5/6*1/2*1/6+till infinity
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Post by indophile Thu Jul 07, 2011 4:32 pm

PseudoIntellectual wrote:Probabilty of her winning 2/7?

1/6+5/6*1/2*1/6+5/6*1/2*5/6*1/2*1/6+till infinity

Perfect!

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Post by indophile Thu Jul 07, 2011 4:35 pm

Question # 2. PI got it.

The answer is 2/7. The probability that Anna wins is the probability that she wins on her first turn or her second turn or her third turn, and so on. The probability distribution, in which we observe a string of failures before the first success is called the geometric probability distribution. The probability that Anna wins on the first roll is 1/6. For Anna to win on the second roll, she must roll any number except 6 on her first roll and Bob must toss tails on his first toss. The probability that Anna wins on the second roll is (5/6)(1/2)(1/6). For Anna to win on the third roll, she and Bob must each have two failures before Anna’s success: (5/6 * ½)^2 * 1/6. So we need to find the sum of infinite geometric series:

P(Anna wins) = 1/6 + (5/6 * ½) * 1/6 + (5/6 * ½)^2 * 1/6 + (5/6 * ½)^3 * 1/6 + ---

We use the formula for the sum of an infinite geometric series, S = a/(1-r), as follows:

P (Anna wins) = (1/6) / (1 - 5/12) = 2/7


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Post by Hellsangel Thu Jul 07, 2011 4:58 pm

indophile wrote:Question # 2. PI got it.

. The probability that Anna wins on the first roll is 1/6

Isn't it 1/12? Probability she rolls a 6 AND Bob lands Tails.
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Post by Mosquito Thu Jul 07, 2011 5:07 pm

Hellsangel wrote:
indophile wrote:Question # 2. PI got it.

. The probability that Anna wins on the first roll is 1/6

Isn't it 1/12? Probability she rolls a 6 AND Bob lands Tails.

If she rolls a 6 she wins. Bob doesnt have to toss.
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Post by Hellsangel Thu Jul 07, 2011 5:29 pm

PseudoIntellectual wrote:

If she rolls a 6 she wins. Bob doesnt have to toss.
OK. I was interpreting her rolling a 6 AND Bob landing heads to be a draw; and both having to play before you could declare a winner or a draw.
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